已知函数f(x)=-x^3+ax^2+1（a∈R)

题目

已知函数f(x)=-x^3+ax^2+1（a∈R)
（1）函数y=f(x)在区间（0,2/3)上递增,在x>=2/3上递减,求a的值
在（1）的条件下,是否存在实数m,使得函数g(x)=x^4-5x^3+(2-m)x^2+1（m∈R)的图像与函数y=f(x)的图像有三个交点,若存在,求出m,若不存在,说明理由

答案

f(x) )=-x^3+ax^2+1
f'(x) = -3x^2 + 2ax =0
x(-3x+2a)=0
x=0 or x= 2a/3
函数y=f(x)在区间（0,2/3)上递增,在x>=2/3上递减
=> 2a/3 = 2/3
a = 1
g(x)= x^4-5x^3+(2-m)x^2+1
f(x) = -x^3+x^2+1 ( a=1)
g(x) = f(x)
x^4-5x^3+(2-m)x^2+1 = -x^3+x^2+1
x^4-4x^3+ (1-m)x^2 = 0
x^2(x^2-4x+(1-m)) = 0
△ of x^2-4x+(1-m) =0
△≥0
=> 16 - 4(1-m)≥0
12+4m≥0
m≥ -3